Answer to Question #285094 in Mechanical Engineering for Johnny

Question

A motor vehicle of mass 2000 kg travels round a curved path of radius 55 m. the

coefficient of friction between the tyres and the road surface is 0.75. The track of the

vehicle is 1.5 m and the centre of gravity is 70 cm above ground level. Calculate the

maximum speed at which the vehicle can travel round the curve:

i. without skidding outwards

ii. without overturning

iii. if the angle of tilt is 300 without overturning

Answer 1

Given:-

M= 2000 kg 

R= 55 m

μ= 0.75

g= 10 m/s² (acceleration due to gravity)

d= 1.5 m (width of the car)

a= 0.7 m (height of the COM from ground)

If the car ain't skidding, that means the car is at equilibrium in the radial direction.

And there are two forces acting in radial direction:-

f= friction = μN = μmg = 0.75×2000×10 = 15000 N (radially inward)

F_c= centrifugal force= mv²/R = 2000×v²/55 (radially outward)

Hence according to equilibrium condition:-

15000= 2000×v^2/55

=> v= (55×15000/2000)^1/2 = 20.31 m/s.  (Ans)

 

 

If the car ain't overturning, then the car is in rotational equilibrium.

There are 2 main forces acting on the car that can possibly rotate the car:-

W= mg= 2000×10= 20000 N (sense- anticlockwise direction)

F_c = Fc= centrifugal force= mv2/R = 2000×v^2/55 (sense- clockwise direction)

The friction will act as a pivoting force on the respective wheels! Hence the wheels act as a pivoted axis or rotational axis.

The weight is acting exactly at the COM (in vertically downward direction) while the centrifugal force is also acting at the COM (but in Horizontal direction)

Lets calculate torque for both the forces:-

τ_mg= mg×(d/2) = 20000×1.5/2 = 15000 Nm

τ_Fc= F_c×a = 2000×0.7×v^2/55

Equating both the torques we get:-

15000= 2000×0.7×v^2/55

v= 24.275 m/s.  (Ans)

 

 

Given:-

θ= 30°. (angle of the slant road)

Now as the car is on this slant road the weight will be splited into 2 components:-

W_x= -mgsinθ. (in the direction parallel to surface)

W_y= -mgcosθ. (in the direction perpendicular to surface)

Same goes for centrifugal force:-

F_cx= mv²cosθ/R

F_cy= - mv²sinθ/R

So the net force in the x direction that will cause a torque of clockwise rotation:-

F_x= mv²cosθ/R - mgsinθ

Hence:-

τ_x= (mv²cosθ/R - mgsinθ)×a

So the net force in the y direction that will cause a torque of anticlockwise rotation:-

F_y= (mv²sinθ/R + mgcosθ)d/2

Equating both the tourques for rotational equilibrium we get:-

(mv²cosθ/R - mgsinθ)×a = (mv²sinθ/R + mgcosθ)d/2

Solving the above equation we get:-

v= 48.76 m/s.  (Ans)